Équations trigonométriques#
Nom, prénom:
Note:
| Question | 1 | Total |
|---|---|---|
| Points | 24 | 24 |
| Obtenus |
Détails des calculs obligatoires. Attention au soin. Calculatrice non
autorisée.
Réponse sous forme de valeur exacte simplifiée.
Question 1 (24 pts)#
Résolvez les équations suivantes (solutions exactes).
\(\cos(\alpha) = -\dfrac{\sqrt{3}}{2}\)
\(\tan(\alpha + 60^\circ) - 1 = 0\)
\(\sin(3x) = \sin(\dfrac{\pi}{3} + x)\)
\(\tan(2x - \dfrac{\pi}{4}) = \sqrt{3}\)
\(\sin(x) - \cos(3x) = 0\)
\(2\cos^2(x) = 1 + \sin(x)\)
Solution
\(\cos(\alpha) = -\dfrac{\sqrt{3}}{2}\)
\(\arccos(\cos(\alpha)) = \arccos(-\dfrac{\sqrt{3}}{2})\)\(\alpha_1 = 150^\circ + k \cdot 360^\circ\)
\(\alpha_2 = -150^\circ + k \cdot 360^\circ\)
\(S = \{150^\circ + k \cdot 360^\circ | k \in \mathbb{Z}\} \cup \{-150^\circ + k \cdot 360^\circ | k \in \mathbb{Z}\}\)
\(\tan(\alpha + 60^\circ) - 1 = 0\)
\(\tan(\alpha + 60^\circ) = 1\)
\[\begin{split}\arctan(\tan(\alpha + 60^\circ)) &= \arctan(1)\\ \alpha + 60^\circ &= 45^\circ + k \cdot 180^\circ\\ \alpha &= -15^\circ + k \cdot 180^\circ \end{split}\]\(S = \{-15^\circ + k \cdot 180^\circ | k \in \mathbb{Z}\}\)
\(\sin(3x) = \sin(\dfrac{\pi}{3} + x)\)
\(\arcsin(\sin(3x)) = \arcsin(\sin(\dfrac{\pi}{3} + x))\)\[\begin{split}3x_1 &= \dfrac{\pi}{3} + x_1 + k \cdot 2\pi\\ 2x_1 &= \dfrac{\pi}{3} + k \cdot 2\pi\\ x_1 &= \dfrac{\pi}{6} + k \cdot \pi \end{split}\]\[\begin{split}3x_2 &= \pi - (\dfrac{\pi}{3} + x_2) + k \cdot 2\pi\\ 3x_2 &= \pi - \dfrac{\pi}{3} - x_2 + k \cdot 2\pi\\ 4x_2 &= \dfrac{2\pi}{3} + k \cdot 2\pi\\ x_2 &= \dfrac{\pi}{6} + k \cdot \dfrac{\pi}{2}\\ \end{split}\]\(S = \{\dfrac{\pi}{6} + k \cdot \dfrac{\pi}{2} | k \in \mathbb{Z}\}\)
\(\tan(2x - \dfrac{\pi}{4}) = \sqrt{3}\)
\[\begin{split}\arctan(\tan(2x - \dfrac{\pi}{4})) &= \arctan(\sqrt{3})\\ 2x - \dfrac{\pi}{4} &= \dfrac{\pi}{3} + k \cdot \pi\\ 2x &= \dfrac{\pi}{4} + \dfrac{\pi}{3} + k \cdot \pi\\ 2x &= \dfrac{7\pi}{12} + k \cdot \pi\\ x &= \dfrac{7\pi}{24} + k \cdot \dfrac{\pi}{2}\\ \end{split}\]\(S = \{\dfrac{7\pi}{24} + k \cdot \dfrac{\pi}{2} | k \in \mathbb{Z}\}\)
\(\sin(x) - \cos(3x) = 0\)
\(\sin(x) = \cos(3x)\)
\(\cos(\dfrac{\pi}{2}-x) = \cos(3x)\)
\(\arccos(\cos(\dfrac{\pi}{2}-x)) = \arccos(\cos(3x))\)\[\begin{split}\dfrac{\pi}{2}-x_1 &= 3x_1 + k \cdot 2\pi\\ -4x_1 &= -\dfrac{\pi}{2} + k \cdot 2\pi\\ x_1 &= \dfrac{\pi}{8} + k \cdot \dfrac{\pi}{2} \end{split}\]\[\begin{split}\dfrac{\pi}{2}-x_2 &= -3x_2 + k \cdot 2\pi\\\ 2x_2 &= - \dfrac{\pi}{2} + k \cdot 2\pi\\ x_2 &= - \dfrac{\pi}{4} + k \cdot \pi \end{split}\]\(S = \{\dfrac{\pi}{8} + k \cdot \dfrac{\pi}{2} | k \in \mathbb{Z}\} \cup \{-\dfrac{\pi}{4} + k \cdot \pi | k \in \mathbb{Z}\}\)
\(2\cos^2(x) = 1 + \sin(x) \qquad \qquad \qquad\) avec \(\cos^2(x) = 1 - \sin^2(x)\)
\(2(1- \sin^2(x)) = 1 + \sin(x)\)
\(2- 2\sin^2(x) = 1 + \sin(x)\)
\(0 = 2\sin^2(x) + \sin(x) - 1\)
On pose \(y = \sin(x)\)
\(2y^2 + y - 1 = 0 \implies (2y^2 + 2y - y - 1) = 0 \implies 2y(y + 1) - (y + 1) = 0 \implies (2y -1)(y + 1)=0\)\[\begin{split}2y -1 &= 0\\ 2\sin(x) -1 &= 0\\ \sin(x) &= \dfrac{1}{2}\\ x_1 &= \dfrac{\pi}{6} + k \cdot 2\pi \quad \text{ou}\\ x_2 &= \pi - \dfrac{\pi}{6} + k \cdot 2\pi \\ &= \dfrac{5\pi}{6} + k \cdot 2\pi \end{split}\]\[\begin{split}y + 1 &= 0\\\ \sin(x) + 1 &= 0\\ \sin(x) &= -1\\ x_3 &= -\dfrac{\pi}{2} + k \cdot 2\pi \end{split}\]\(S = \{\dfrac{\pi}{6} + k \cdot 2\pi | k \in \mathbb{Z}\} \cup \{\dfrac{5\pi}{6} + k \cdot 2\pi | k \in \mathbb{Z}\} \cup \{-\dfrac{\pi}{2} + k \cdot 2\pi | k \in \mathbb{Z}\}\)